Misconceptions about the nature of electricity, as taught from school books.

[k4kyv]

How does electricity work? If you've learnt about the nature of electricity from grade-school (or even high school or university) textbooks, you may need some "debunking" to find out how electricity DOESN'T work.

http://amasci.com/miscon/elect.html (links)

http://amasci.com/miscon/eleca.html#frkel (text file)



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[K8ERV]

How does electricity work?


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I don't know. I don't even know how I work. If at all.

TOM K8ERV Montrose Colo

[K1CJS]

What a shocking subject!

On the serious side, however, I find myself agreeing with several of the suppositions made by this man, and disagreeing with others. On that note, I saw no references to his educational experiences other than practical experience as a research engineer. I suppose that does give him somewhat of an edge on the rest of us--but how many of us also have the 'practical experience' of working with electricity and electronics every day?

So, I suppose this article maybe considered 'food for thought'.

[ky5u]

Load of semantic crap. I love the whole "debunking" movement. Some idiot with his butt up in the air because HE can correct "accepted knowledge". It's all about insecurity and "size". Sigmund, where are you?

[N2RJ]

I especially love how he claims that AC makes electrons only "wiggle" and not actually move. :D

[ve2nsm]

Some of the stuff is disturbing, not that I completely disagree but instead of making things clear it's becoming more muddled.

I think most of his peeves are based on the wrong usage of words, he's kind of picky.

[wa9cwx]

I DID understand simple electrical circuits for 45 years, NOW I understand nothing.......:D

Give me a break. :p

It is like telling a mechanic that there IS no internal combustion action.
It is, instead, an expansion of super heated gas that increases the distance between molecules that have been oxygenated quickly.....

He obviously has taken a common concept, and made a point of showing how real scientific investigation has led to a different detailed analysis than the popular one......

NO S**T Sherlock !!!

EVERY 'Popular' (Grammar school) explanation of ANYTHING is flawed, simplified, and short circuited for easy consumption.

Anyone who gets through High School should have figured THAT out.

The very nature of teaching BASICS about ANY concept that is a part of advanced study INCLUDES simplification, and the resultant 'error' that results from that.

SHEEESH !

[ke7tfn]

Bumble Bees can not fly!!!!!

Lets see, I learned about electronics in the Navy, while at the same time the opposite theories were being taught in the colleges.

HOWEVER did you know that the electronics on Naval ships worked just fine!

And the electronics in the civilian sector worked just fine?

How can this be?

LOL if you want to come up with something neat and new, make sure that it is not 80 years old.

[KI6NNO]

That entire group of links
http://amasci.com/miscon/elect.html (links)
is mostly crap. Seriously, crap, crap crap. :eek: :eek: :eek:

For example:
When you connect a light bulb to a battery, Electrical Energy moves from the battery to the bulb. This is a one-way flow.
No, if that were the case, two wires wouldn't be necessary.

The battery loses energy and the bulb gains it.
Not really - the bulb dissipates the electrical energy as soon as it "gains" it. This is work, not simply energy (See the First Law of Thermodynamics below) because entropy is not transferred with the electrical energy. (with entropy being "a system's unavailability to do work" <-- paraphrased). Thus there is no gain of energy into the lightbulb as incoming energy is immediately converted to physical changes in the filament (from heating) and the transfer of heat via conduction, convection, and radiation to the surrounding space (or atmosphere, it doesn't really matter what surrounds - or doesn't surround - the lightbulb).

Then the energy received by the bulb is turned into light. If this phenomenon is examined in great detail, we find that electrical energy is composed of waves traveling along columns of electrons inside the wires, and the energy itself is contained in electromagnetic fields connected to those electrons. We find that it travels as wave energy, that it exists only outside of the wires...
No. While AC electron flow and the resulting EM fields occurs in "waves", DC electron flow does not, although there is still an EM field and energy is stored in that field. "Skin effect" determines how deep the electron flow penetration is for AC flow, while DC flow penetrates the entire conductor. Furthermore, the EM field outside a conductor, while it does store energy, contributes nothing to the work done on the lightbulb until the EM field collapses. However, this collapse does not occur with DC current flow thus the energy being converted to work in the lightbulb must be getting there from some other means: the electrons themselves. This is also true for when an EM field collapses in AC current flow or when DC flow is interrupted. The EM field collapse "pushes" electrons in one way or the other depending on the EM field polarity, and additional work is done on the system - again by moving electrons.

and most importantly, that it TRAVELS ONE WAY ALONG BOTH WIRES on its trip from the battery to the bulb.
No. While this could be interpreted as true for AC for the phase component that "pushes" electrons into the bulb" from each side of the bulb during the appropriate phase, it is definitely not true for DC current flow. Ohm's law clearly shows that the relative "energy" contained in the electrons "leaving" the lightbulb is less than the energy of the electrons "entering" the lightbulb - thus the drop in voltage potential across the lightbulb in both directions for AC flow and in a single direction for DC flow. However, the electrons leaving the lightbulb are not at Absolute Zero* and therefore retain both kinetic and potential energy.

*Even at absolute zero there is electron motion within atoms - the electrons don't just "fall out of the space surrounding the nucleus", thus there is still energy in each atom and electron cloud surrounding atomic nuclei at absolute zero. Not that we can cause an specific object to reach and stay at Absolute Zero by any finite amount of processes...

The First Law of Thermodynamics clearly states that the reason a battery or generator "loses" energy is because of the work done on the lightbulb. Work is not energy thus the lightbulb never gains energy.
The electrical energy did not travel in a circle. So, when you plug a lamp into a wall socket, you shouldn't imagine that the AC energy is a mysterious invisible entity traveling back and forth inside the wires.
Yes it did, although it's not mysterious at all. The voltage potential "pushes" (or drags - however you want to look at it) electrons from one atom to another - like a bucket brigade. For DC , the "buckets" go in one direction only, for AC, they go both ways, but only one way at a time.

Instead you should think of AC energy as a mysterious invisible flow that comes out of the outlet, runs along the outside of BOTH wires, then dives into the filament of the light bulb. Your electric company is sending out long "sausages" of electrical energy, the wires are guiding them, and your appliances are absorbing them.

Maybe for RF, but not 60Hz and certainly not for DC. If that were the case, the power company would surely use hollow conductors, as would virtually every other manufacturer in the world, erm, universe (gotta include E.T. here).
Skin effect is in the General Exam pool, and the formula for determining electron flow depth into a conductor is clear and well founded. I would expect General licensees to at least question the poster's premise on this topic.

The vast majority of what's posted in there further abuses physics and thermodynamics, and fails miserably against syllogistic logic - unfortunately to the detriment of lay people who might not know enough of the appropriate sciences to recognize fact from fiction because the fiction is an interesting, if incorrect, read.

73 de Dave KI6NNO

[K1CJS]

I DID understand simple electrical circuits for 45 years, NOW I understand nothing.......:D

Give me a break. :p

It is like telling a mechanic that there IS no internal combustion action.
It is, instead, an expansion of super heated gas that increases the distance between molecules that have been oxygenated quickly.....

He obviously has taken a common concept, and made a point of showing how real scientific investigation has led to a different detailed analysis than the popular one......

NO S**T Sherlock !!!

EVERY 'Popular' (Grammar school) explanation of ANYTHING is flawed, simplified, and short circuited for easy consumption.

Anyone who gets through High School should have figured THAT out.

The very nature of teaching BASICS about ANY concept that is a part of advanced study INCLUDES simplification, and the resultant 'error' that results from that.

SHEEESH !

Well, if you put it that way, it instantly clearer! As Col. Sherman Potter (MASH) would say (to Beaty's article)---"Horsehockey!" :)

[ab0wr]

I love the part about "free" electrons existing in a metal with no charge flow through it. I guess they just "hang around".

As ki6nno notes, it takes energy to knock an electron out of its "statistical location cloud", i.e. its "orbit". They just don't "float" away.

His spiel about electric "current" is also somewhat mind numbing. Current in amperes is defined in terms of coloumbs/seconds or "electric charge" per second. So current *is* the flow of electric charge. To say that the term current is being misused and that charge flow should be used instead requires a viewpoint I don't understand.

tim ab0wr

[N2RJ]

I love the part about "free" electrons existing in a metal with no charge flow through it. I guess they just "hang around".


That is actually high school chemistry and is correct:

http://en.wikipedia.org/wiki/Metallic_bond

Metal atoms contain few electrons in their valence shells relative to their periods or energy levels. Such electrons can stray easily from the atoms and become delocalized, forming a sea of electrons permeating a giant lattice of positive ions. The freedom of conduction electrons to migrate gives metal atoms, or layers of them, the capacity to slide past each other, giving rise to metals' typical characteristic phenomena of malleability and ductility.

http://www.britannica.com/eb/article-9052274/metallic-bond

Such a solid consists of closely packed atoms. In most cases, the outermost electron shell of each of the metal atoms overlaps with a large number of neighbouring atoms. As a consequence, the valence electrons continually move from one atom to another and are not associated with any specific pair of atoms.

[k0cmh]

Electricity has been known for many years. It is all rather simple, and it seems that the authors of the various articles are playing word games, and themselves furthering "wives tales" of their own.

1. AC is nothing more than DC in a constant state of changing voltage.

2. DC current moves through a metallic conductor by collisions of electrons. The EMF causes on electron to "jump" through the conductor until it collides with another electron. That electron get "bumped" along unitl it collides with another , etc. etc. Physicists have calculated the actual linerar movement of individual electrons in DC current as very small and slow. Individual electrons in DC currnet flow move in a metallic conductor in the area of inches per hour. It takes a lot of collisions for them to move much in our world of measurements.

3. There IS a "sea" of electrons availabel for current movement in a metal conductor, but most of the electrons are "closely bound" to the atom nucleus. The "sea" of electrons are those being "shared", which are loosly held by the nucleus. These are what are responsible for the current in a metal conductor. They are much more easily moved than those electrons in orbitals "innermost" to the nucleus. It is all a matter of how much energy it takes to bump an electron. The unique structure of metallic crystals causes a lot of very loosly held ("shared") electrons availble for easy movement.

[AE6IP]

Yup. In addition, the uncertainty principle applies, and it's the reason quantum tunnelling works, and also Hawking radiation, which allows energy to escape the event horizon of a black hole.

God may not play dice with the universe, but Heisenberg certainly did. ;)

[WA0LYK]

That is actually high school chemistry and is correct:

http://en.wikipedia.org/wiki/Metallic_bond

Metal atoms contain few electrons in their valence shells relative to their periods or energy levels. Such electrons can stray easily from the atoms and become delocalized, forming a sea of electrons permeating a giant lattice of positive ions. The freedom of conduction electrons to migrate gives metal atoms, or layers of them, the capacity to slide past each other, giving rise to metals' typical characteristic phenomena of malleability and ductility.

http://www.britannica.com/eb/article-9052274/metallic-bond

Such a solid consists of closely packed atoms. In most cases, the outermost electron shell of each of the metal atoms overlaps with a large number of neighbouring atoms. As a consequence, the valence electrons continually move from one atom to another and are not associated with any specific pair of atoms.

First, don't believe everything you read on wikipedia. If electrons could become "delocalized" and form a "sea of electrons", then you could stand a metal bar on end and see the "free" electrons run out of the end of it! (A positive ion means the electron is gone from that atom and is roaming free.) This doesn't happen because they are always localized, even if they are constantly moving from locality to locality (between atoms).

The second definition is much more accurate. The electrons in the outer valence shells overlap each other and individual electrons end up being "shared" between many atoms. This is far from being "delocalized" or having "free" electrons floating around without a corresponding atom.

I guess you could make the argument that delocalized means shared between atoms, but this isn't exactly what the wiki definition is describing because they use the term positive ion, i.e. missing an electron.

Jim
WA0LYK

[KA4DPO]

Everyone knows electricity comes from carpet. :D

[N2RJ]

First, don't believe everything you read on wikipedia. If electrons could become "delocalized" and form a "sea of electrons", then you could stand a metal bar on end and see the "free" electrons run out of the end of it! (A positive ion means the electron is gone from that atom and is roaming free.) This doesn't happen because they are always localized, even if they are constantly moving from locality to locality (between atoms).

The second definition is much more accurate. The electrons in the outer valence shells overlap each other and individual electrons end up being "shared" between many atoms. This is far from being "delocalized" or having "free" electrons floating around without a corresponding atom.

I guess you could make the argument that delocalized means shared between atoms, but this isn't exactly what the wiki definition is describing because they use the term positive ion, i.e. missing an electron.

Jim
WA0LYK

In school I was taught about the giant sea of mobile electrons. That's why I quoted Britannica as well, because I knew someone would say, "but, but, but... Wikipedia!!!"

[KM5FL]

Everyone knows electricity comes from carpet. :D

And car seats.. :D


KM5FL

[wa3vjb]

It is like telling a mechanic that there IS no internal combustion action.
It is, instead, an expansion of super heated gas that increases the distance between molecules that have been oxygenated quickly.....


Oh no you don't.
I have a fundamental understanding of THIS situation, and you must accept the following:

The engine in a car takes the movement of the accelerator and amplifies it, transferring that power to revolve the wheels.

This has been proven across many engines and over a long period of time.

http://www.wa3vjb.com/pics/IMG_0815'.JPG

[ab0wr]

That is actually high school chemistry and is correct:

http://en.wikipedia.org/wiki/Metallic_bond

Metal atoms contain few electrons in their valence shells relative to their periods or energy levels. Such electrons can stray easily from the atoms and become delocalized, forming a sea of electrons permeating a giant lattice of positive ions. The freedom of conduction electrons to migrate gives metal atoms, or layers of them, the capacity to slide past each other, giving rise to metals' typical characteristic phenomena of malleability and ductility.

http://www.britannica.com/eb/article-9052274/metallic-bond

Such a solid consists of closely packed atoms. In most cases, the outermost electron shell of each of the metal atoms overlaps with a large number of neighbouring atoms. As a consequence, the valence electrons continually move from one atom to another and are not associated with any specific pair of atoms.

Actually I reread the article and found that he and I seem to agree more than disagree. Where he talks about charge flow he talks about it *inside* the metal.

Free electrons outside the metal would create an electrcostatic charge between the surrounding electron cloud and the metal itself. This would create an electric potential that could actually be measured. I've never been able to hold a piece of copper wire up to an electroscope and get a reading.

Having a "cloud" of electron *outside* the metal is vastly different than having electrons able to freely migrate between atoms *inside* the metal or even in having positive and negative ions *inside* the metal itself.

Gauss' Law states:" If the volume within an arbitrary closed mathematical surface holds a net electric charge Q, then the electric flux [Phi] though its surface is Q/[epsilon]0 " (that's epsilon sub-zero, the permittivity of free space). If you had a cloud of free electrons surrounding a conductor there would be an electric flux between that "cloud" and the conductor itself since you could define a cylindrical surface that would exist outside the conductor itself and inside at least part of the "cloud" of electrons. I've never actually seen that demonstrated to be the case. Does someone have a reference that shows that this can be demonstrated or measured?

tim ab0wr

[KI6NNO]

Free electrons outside the metal would create an electrcostatic (sic) charge between the surrounding electron cloud and the metal itself. This would create an electric potential that could actually be measured. I've never been able to hold a piece of copper wire up to an electroscope and get a reading.

You can't because it won't occur without something else causing a charge to be applied to the wire. Also, wire, because of it's nature, can't hold a static charge on it's own thus there's nothing differential to measure in that case.

Having a "cloud" of electron *outside* the metal is vastly different than having electrons able to freely migrate between atoms *inside* the metal or even in having positive and negative ions *inside* the metal itself.
This is usually a bad thing (i.e. corona) and one of the reasons that conductors are insulated - to prevent electrical charge moving outside of the conductor.

...If you had a cloud of free electrons surrounding a conductor there would be an electric flux between that "cloud" and the conductor itself since you could define a cylindrical surface that would exist outside the conductor itself and inside at least part of the "cloud" of electrons. I've never actually seen that demonstrated to be the case. Does someone have a reference that shows that this can be demonstrated or measured?

tim ab0wr

Ummm... that's a Corona, generally speaking. You can see this inside tubes, and sometimes in tube glass, and sometimes see or hear it near power lines (not to be confused with a plasma). Yes, there's a flux, because the charge is radiated away in the EM field as planar waves (Maxwell) because the particles themselves are energetic (Planck, et. al.).

This leads back to not being able to measure a charge in a wire. Since any electrons that would become energetic enough to register a "charge" on an electroscope would dissipate the charge as radiation immediately, there's no differential charge in the wire to measure. This also goes back to thermodynamics regarding conservation of charge, so without something external adding or taking charge away from the wire, there's nothing differential to measure and thus no reading on an electroscope.

There is no "cloud of electrons outside of a conductor" unless you put enough potential into the conductor that it causes ionization external to the conductor (think static electricity).

It's also not so much the electrons that are mobile over macro-scale distances within a metal (conductor), it's the electrical charge that's mobile. This can be visualized in an ocean or earthquake wave: While an ocean wave (the "energy") can travel 600mph across an open sea, the water molecules that make up the wave actually travel very little from their original location in the ocean - they more or less move in a circle. Seismic "P" waves are more longitudinal rather than up-and-down and more closely resemble charge moving through a conductor. In this example, the energy is transmitted through the earth by molecules moving up to a few feet back and forth, but the information (the energy) is transmitted a long distance.

There's actually an "aha" moment here because the information in a wave (or wire) can't travel faster than the cosmic speed limit (the C in E=MC^2) for whatever is moving in that medium. While mechanical systems transfer information (energy) relatively slowly because the particles are massive, electromagnetic systems can transfer at near-speed of light speeds because electrons in metal are quite mobile over nuclear-scale distances, very light so they can be quickly accelerated (moved from their shell), and the charge is fungible (indistinguishable from one electron to another). Thus when a charge is applied to one end of a conductor, an equivalent charge appears at the other end at nearly the speed of light, whilst the individual electrons in the conductor barely moved at all, and on a macro scale the electron shells of the metal atoms remain filled. Pretty cool, imo. :cool:

Anyway, I'm babbling on about the latter stuff, I just find it interesting and thought it was somewhat relevant to the topic.

73, Dave

[KD0DKI]

Computers are magic.

Just like religon.

[K0FK]

I liked the part about the sausages.

[K1CJS]

The more I read the article, the more I see that the author has his own esoteric ideas about simple everyday physics. I'd hate to see what his explanations about advanced physics.

Unfortunately, the internet allows everybody to become a writer and a publisher--and instead of promoting greater understanding, is actually promoting misrepresentations of generally accepted ideas and methodology.

If you need or want information about concepts and ideas, why not go to an accepted knowledge website instead of going to sites such as wikipedia and read publisher rejected materials such as the one this article is all about.

[KI4NGN]

The more I read the article, the more I see that the author has his own esoteric ideas about simple everyday physics. I'd hate to see what his explanations about advanced physics.

Unfortunately, the internet allows everybody to become a writer and a publisher--and instead of promoting greater understanding, is actually promoting misrepresentations of generally accepted ideas and methodology.

If you need or want information about concepts and ideas, why not go to an accepted knowledge website instead of going to sites such as wikipedia and read publisher rejected materials such as the one this article is all about.

I couldn't agree with you more.

I read more and more quotes from wiki's stated as if they're gospel, people forgetting where the content of the wiki's comes from.

Mike

[ab0wr]

You can't because it won't occur without something else causing a charge to be applied to the wire. Also, wire, because of it's nature, can't hold a static charge on it's own thus there's nothing differential to measure in that case.

Yep, that's exactly what I was trying to say.

This is usually a bad thing (i.e. corona) and one of the reasons that conductors are insulated - to prevent electrical charge moving outside of the conductor.
As you note below, corona is actually ionization of the air surrounding something with a high voltage potential compared to its surrounding environment. In this case the insulation is actually protecting the *air* instead of humans!

Ummm... that's a Corona, generally speaking. You can see this inside tubes, and sometimes in tube glass, and sometimes see or hear it near power lines (not to be confused with a plasma). Yes, there's a flux, because the charge is radiated away in the EM field as planar waves (Maxwell) because the particles themselves are energetic (Planck, et. al.).

There is a flux because of the creation of charged particles in the air itself not because of a "cloud" of free electrons from the conductor itself.

I don't believe electric flux lines between charged particles are the same thing as radiated electromagnetic waves with an E and an H field. An electroscope will detect electric flux between charged particles but it won't detect the radiated field from an antenna, at least not any antenna I've ever been around. Charged particles *do* emit radiation but I was taught that this was in the form of photons emitted as the charged particles fall back to lower exited states. I was taught in physics that the photon has no charge. Has this changed in th last 40 years?

This leads back to not being able to measure a charge in a wire. Since any electrons that would become energetic enough to register a "charge" on an electroscope would dissipate the charge as radiation immediately, there's no differential charge in the wire to measure. This also goes back to thermodynamics regarding conservation of charge, so without something external adding or taking charge away from the wire, there's nothing differential to measure and thus no reading on an electroscope.
More eloquently put than I could do it.

There is no "cloud of electrons outside of a conductor" unless you put enough potential into the conductor that it causes ionization external to the conductor (think static electricity).
Yep, ionize the air itself.

There's actually an "aha" moment here because the information in a wave (or wire) can't travel faster than the cosmic speed limit (the C in E=MC^2) for whatever is moving in that medium. While mechanical systems transfer information (energy) relatively slowly because the particles are massive, electromagnetic systems can transfer at near-speed of light speeds because electrons in metal are quite mobile over nuclear-scale distances, very light so they can be quickly accelerated (moved from their shell), and the charge is fungible (indistinguishable from one electron to another). Thus when a charge is applied to one end of a conductor, an equivalent charge appears at the other end at nearly the speed of light, whilst the individual electrons in the conductor barely moved at all, and on a macro scale the electron shells of the metal atoms remain filled. Pretty cool, imo. :cool:

This is where some of his explanations are a little off I think. The actual migration speed of any specific electron in a conductor may not be all that fast but the actual speed of transmission doesn't depend on the migration speed but on the speed of inserting charge and extracting charge.

tim ab0wr

[KI6NNO]

There is a flux because of the creation of charged particles in the air itself not because of a "cloud" of free electrons from the conductor itself.

I don't believe electric flux lines between charged particles are the same thing as radiated electromagnetic waves with an E and an H field. An electroscope will detect electric flux between charged particles but it won't detect the radiated field from an antenna, at least not any antenna I've ever been around.

Hi Tim,
I'm going to have to take a moment on this, and *I think* touch on some Heisenberg principle and calculus (sorry!). Just skip to "Application" for the short(er) answer.

Description:
An electroscope is a destructive analysis tool in the sense that the Heisenberg priciple (a take on the "no free lunch" rule) requires that a system is perturbed in order to take a measurement of the system. In the case of charge, some amount of electrical charge must be transferred to the electrode leaves in the electroscope, thus reducing the amount of charge originally present within the system. You'll also notice that once the electroscope is removed from the system it's measuring, the leaves will stop being repelled from each other. This is due to the charge "leaking out" of the electroscope as the charge on the leaves finds its way through the sensing electrode back out into a lower charge potental (and/or the polarization between the sensing electrode tip and the leaves equalizing - it depends on how you look at it). That lower potential can be the atmosphere, a lower charged "E-flux line" thus returning the sensed charge back to the measured system, whatever. There are some differences between "free" (electrical polarization) and "bound" (nuclear polarization) charge relating to a charged object, but for simplicity's sake the electroscope is working on the free charge from an object. Ok, enough on that, I think.

The E-flux itself is the integral (summation of) of the vector (directional value of the) electrical field times the surface area of the electrical field being measured. Unfortunately, I don't know how to post the formula, but here's a wiki page:
http://en.wikipedia.org/wiki/Flux

In a DC or electrostatic system, the E (electrical) and H (magnetic) fields are each created in one vector (direction) only, thus when an electroscope probe is placed in a DC or electrostatic E field, a constant (but now slightly less than what was there originally [Heisenberg]) charge is coupled to the electroscope leaves and the leaves repel each other.

In an AC system, the electroscope couples charge to the leaves exactly the same as in a DC/electrostatic system. However, the E and H fields vary over time, thus the charge coupled into the electroscope varies over time. This can be visualized by using a AC frequency of say 0.5 Hz and watching the electroscope leaves repel and then return to their resting position each half-cycle. For 1/2 the cycle, the leaves will be positively charged, and for the other half of the cycle, negatively charged (and interestingly enough the tip of the sensing electrode is oppositely charged from the leaves unless the electrode is directly connected to the object being measured). :cool:

Application:
As AC frequency increases, the electroscope simply isn't able to respond to the changing E field fast enough due to the mass of the leaves and the leaves won't move from their resting (discharged) position. However, it is possible to add a diode-pump (say to collect charge from the positive phase of the E field) to the electroscope sensor and prove that there is an e-field present. (An electroscope is essentially a capacitor and will behave exactly like its cousin the Leyden jar in storing a charge).

Another thing that complicates this is that a collapsing E field propagates an expanding H field and vice-versa, thus the original charge is radiated away in an AC system at any frequency (Maxwell).


Charged particles *do* emit radiation but I was taught that this was in the form of photons emitted as the charged particles fall back to lower exited states. I was taught in physics that the photon has no charge.

Has this changed in th last 40 years?

Nothing has changed here. A photon can be defined as both a particle and a wave. It simply depends on how the photon is measured. The photon has no net charge, yet the photon is the carrier "particle" for EM radiation. Here's a wikipedia excerpt:

"...a photon can be considered as a mediator for any type of electromagnetic interactions, including magnetic fields and electrostatic repulsion between like charges."
http://en.wikipedia.org/wiki/Photon

This means that the energy within any frequency EM wave can be contained within a photon, and can be released in any form (in a general sense) when a given photon interacts with another "particle".

So, while the definitions of a photon and the behavior of electrons moving closer to a ground state (and emitting photons) still hold true from 40 years ago, our understanding of the physics behind their behavior has been refined somewhat.

Anyway, I hope this helps.

73, Dave

(fwiw, I like to use Wikipedia as it's easily accessible, although not always as accurate as primary encyclopedic references. ymmv)

[KB4QAA]

Wikipedia: Where anyone can delete accurate information! ;)

This self purported expert isn't 100% correct in his descriptions either.

Of course K-6 textbooks aren't 100% accurate and detailed. They are simplifications intended to describe GENERAL CONCEPTS. At grade 6 they still haven't been introduced to algebra and couldn't even solve basic Ohms Law calculations. And this nut expects that Kindergartners should be able to comprehend and regurgitate the differences between electron vs. hole flow theory? bah.

b.

[K8MHZ]

At grade 6 they still haven't been introduced to algebra and couldn't even solve basic Ohms Law calculations.Really?

How do you explain my daughter at age 11 passing her Tech exam with ease?

There are many others that have done it at an even younger age.

I thought the link to be *very* thought provoking and I will spend some more time going over it. I work with electricity everyday and I saw nothing, so far, that was absolutely disputable in the work that was linked. My thanks to the OP for putting it here for us to all discuss.

[KI6NNO]

Wikipedia: Where anyone can delete accurate information! ;)
hehe, that is truly my pet peeve with it!

This self purported expert isn't 100% correct in his descriptions either.
I saw this and thought "uh oh, what did I miss..."
I'm hoping you meant the guy in the link ;)

Of course K-6 textbooks aren't 100% accurate and detailed. They are simplifications intended to describe GENERAL CONCEPTS. ...
Thus the problem with any model, including the stuff I've posted here. It's terribly difficult to explain things in lay terms without losing the detail needed for technical accuracy.

-->> this part deleted because it was rambling off-topic crap hihi. I left the picture though <--

73, Dave

[KB4QAA]

Really?

How do you explain my daughter at age 11 passing her Tech exam with ease?

There are many others that have done it at an even younger age.
.
I'm not saying that 6th graders aren't capable. I'm merely pointing out that algebra is not part of the curriculum at these grade levels, thus, they would not be expected to have any grasp of even how to do Ohms law manipulations. Hence going overboard into minute details of electricity is pointless.

Personally, I feel one of the shortcomings in math education was treating higher level concepts as being "off limits" served to restrict curiosity and enthusiasm for learning more math.

73, Bill

[k4kyv]

Originally Posted by KB4QAA
Wikipedia: Where anyone can delete accurate information!

Only if the moderators allow it to be deleted, not to say that their wisdom is any closer to infinite than that of the moderators of this or other internet message boards. I haven't found Wiki to be particularly better or worse than the other popular encyclopaedias, and its accuracy is probably on par with that of the printed and electronic news media. When in doubt, cross reference with other sources of the same information. Google and other search engines can be useful for that, as well as the public library or your own personal library.

[AF9J]

2. DC current moves through a metallic conductor by collisions of electrons. The EMF causes on electron to "jump" through the conductor until it collides with another electron. That electron get "bumped" along unitl it collides with another , etc. etc. Physicists have calculated the actual linerar movement of individual electrons in DC current as very small and slow. Individual electrons in DC currnet flow move in a metallic conductor in the area of inches per hour. It takes a lot of collisions for them to move much in our world of measurements.


Ummmm, that's not quite true. The cross section for collisions of electrons is quite small (for example: much moreso, than for nuclear fission reactions). So, while an electron can physically collide with another electron, knocking it free to become a free electron, the probability is pretty low. The best way to put it, is that when a free electron enters the electrically neutral system (protons = electrons) of an atom), for the system to remain electrically neutral, it has to emit an electron. Which one leaves the system, is determined more by the probability of each electron, to leave the atom's system (which is where quantum mechanics comes into play). BTW, if you want to get esoteric, by quantum mechanics, electrons are NOT in orbits (as Nils Bohr postulated), but are at specific energy levels, in generalized "clouds" which are defined by the probability of electrons existing the cloud.

Also, why the hype about Heisenberg? Other than postulating his uncertainty principal, he didn't do much else for Quantum Physics. Schrodinger did more in that regards.

73,
Ellen - AF9J
B.S. - Nuclear Engieneering

[W1GUH]

Load of semantic crap. I love the whole "debunking" movement. Some idiot with his butt up in the air because HE can correct "accepted knowledge". It's all about insecurity and "size". Sigmund, where are you?

Of course, we don't have any of those here, do we? :o

All this talk about electrons, and the poor, overworked and overlooked "hole" is completely forgotten. But then, this new-fangled solid-state stuff will never catch on! :D

[KI6NNO]

Also, why the hype about Heisenberg? Other than postulating his uncertainty principal, he didn't do much else for Quantum Physics. Schrodinger did more in that regards.

73,
Ellen - AF9J
B.S. - Nuclear Engieneering

Hi Ellen,

Only to keep a common point in the response about measurement perturbing a system.

I left out the uncertainty principle coming out of Matrix Mechanics which Heisenberg jointly developed and which was very theoretical at the time; determining that electrons could not possibly be constrained to Bohr's classical orbital structure. Both scientists received the Nobel Prize for their work in Physics in the early 30's, and as you noted Schroedinger definitely produced more in the latter parts of his career, leading to further advances. All great stuff!

I couldn't imagine explaining barns with regards to e- e- interactions. Yikes! :eek: :eek: :eek:

Well, maybe I could. I can be a sucker for punishment sometimes. :rolleyes:

73, Dave

[ab0wr]

Originally Posted by ab0wr View Post
I don't believe electric flux lines between charged particles are the same thing as radiated electromagnetic waves with an E and an H field. An electroscope will detect electric flux between charged particles but it won't detect the radiated field from an antenna, at least not any antenna I've ever been around.
Hi Tim,
I'm going to have to take a moment on this, and *I think* touch on some Heisenberg principle and calculus (sorry!). Just skip to "Application" for the short(er) answer.

Dave,

I'm sorry but we must be talking past each other.

Neither DC or AC current generates electric flux lines around a conductor carrying current. They generate a magnetic field (which is how a transformer or electromagnet work).

Electric flux lines can only begin and end on something which has charge.

Again, using Gauss' Law, if you can draw a surface through which flux lines penetrate then there are charges inside and outside the surface.

Charges can migrate to the *surface* of a conductor but not outside the conductor otherwise you could draw a Gauss surface just outside the surface of the conductor and have charges inside and outside the surface generated from the current flowing in the wire.

I've never seen a textbook that shows how this could take place.

It isn't an issue of Heisenberg or calculus. You *can* get electric flux lines between two conductors of different charge (e.g. a capacitor) but the charges that cause the electric flux accumulate on the surfaces of the conductors and not outside the conductors.

tim ab0wr

tim ab0wr

[AF9J]

Hi Ellen,

I couldn't imagine explaining barns with regards to e- e- interactions. Yikes! :eek: :eek: :eek:

Well, maybe I could. I can be a sucker for punishment sometimes. :rolleyes:

73, Dave

LOL!! Yep -the almighty Barn. 1 Barn = 10 to the minus 24th power, square centimeters.

73,
Ellen - AF9J

[KI6NNO]

Dave,
I'm sorry but we must be talking past each other.
tim ab0wr

Hi Tim,
No worries, we probably are. :D

Electric flux lines can only begin and end on something which has charge.
Yes. Gauss' law can even be used to describe the behavior of a single electrical point charge.
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gaulaw.html

Neither DC or AC current generates electric flux lines around a conductor carrying current. They generate a magnetic field (which is how a transformer or electromagnet work).

It depends on your frame of reference (Einstein, Special relativity).

If a conductor has a voltage potential of say 1V with respect to ground then there is a difference in charge between the conductor and ground and an electric flux must exist between those two objects. It doesn't matter if this is a wire vs. another object with a different charge, or two (or more) single point charges, there is always an electric field between two (or more) charged objects.

A stationary charge such as the 1V potential conductor in an open circuit does not have a magnetic field (Maxwell). However a moving charge has both an electric and magnetic field (again, Maxwell). If the magnetic field should vary, such as in an AC circuit, then the E-field will also vary, and the varying E-field will induce a change in the H-field, and so on...

The electric and magnetic fields are both part of the same force as described by special relativity because it depends on the space-time frame of the observer as to how the EM fields are interpreted (Einstein).

Again, using Gauss' Law, if you can draw a surface through which flux lines penetrate then there are charges inside and outside the surface.
Yes.

Charges can migrate to the *surface* of a conductor but not outside the conductor otherwise you could draw a Gauss surface just outside the surface of the conductor and have charges inside and outside the surface generated from the current flowing in the wire.

I've never seen a textbook that shows how this could take place.
If this were true, we wouldn't need insulation.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gausur.html

The thing is, while a Gaussian surface allows us to visualize what's going on in a charged system, no material exists that allows a Gaussian surface to exist in its textbook form, simply because there are no perfect conductors and no perfect dielectrics.

Two of the things left out of this conversation so far are the conductivity and permittivity of the material in which the fields pass through. Copper metal is very conductive while air is much less so. This means that charges will flow easily within the copper conductor and less so in the air outside of it.

When we measure charge in a conductor, we're essentially creating a capacitor with our measuring device so we're also measuring permittivity - in essence the ability of a material to store charge, or in other words, higher permittivity allows for an equivalent charge to be stored at a lower e-field strength.

For an example with a bare copper wire in air, the copper has low permittivity and high conductivity and the air has a relatively high permittivity and low conductivity. This means that the air as a dielectric is able to store more charge over a given surface area than a wire can and the charge is less able to migrate from one potential to another in the air than in the conductor. Thus, the charges in the conductor "collect" on the surface of the conductor closest to the area(s) of different potential that exist across the air gap between the conductor and the measurement device.

However, air and every other material in the universe does have some amount of conductivity and thus some amount of charge will migrate beyond the surface of a conductor into whatever medium is surrounding it. Going back to the point charge example, anytime a difference in charge exists between two points, an electrical field exists between those points as well.

So, because of this, you can draw a (closed) gaussian surface outside of the conductor and still measure all of the flux described from the area "below" the gaussian surface. Gauss' law doesn't say there can't be any charge past the closed surface, it does say that only the charge included "below" the surface is included in the flux coming "out" of the surface.

It isn't an issue of Heisenberg or calculus.
I don't know how to properly discribe these fields without calculus, so I guess you've got me there. Then again, I don't know how anyone else could either because Maxwell's equations use calculus unless one were to break integrals into their algebraic summation equivalents. :confused:

You *can* get electric flux lines between two conductors of different charge (e.g. a capacitor) but the charges that cause the electric flux accumulate on the surfaces of the conductors and not outside the conductors.

For the most part, yes. Again, because all objects are both permittive and conductive, some amount of charge must exist outside of the conductors. As the "electric flux" is the sum of the vectors of the electrical field, those external charges also contribute to the measured value of the electric flux.

This can actually be measured in electrolytic* capacitors that have been charged for a very long time as a "battery effect":

Because of the significant length of time of remaining charged, some of the charge on the surface of the plates (Maxwell's displacement current - which is the non-flowing portion of the total current) "leaks" into the dielectric. If the capacitor is fully discharged and left to rest, the voltage potential across the plates will rise again as the "excess" charge migrates back out of the dielectric and onto the capacitor plates. This would not occur if charges that cause electric flux could not move beyond the surface of a conductor.

*any capacitor will do this, but it's commonly observed in large electrolytics due to the large capacitance involved.

Well, in any event, barring typos or other brain farts, this is my understanding of how this all works. Whew, I'm beat!

73, Dave

[N2RJ]

Really?

How do you explain my daughter at age 11 passing her Tech exam with ease?

Because the tech test is mostly common sense, and the questions and answers (question pool) are available beforehand.

[N2RJ]

Wikipedia: Where anyone can delete accurate information! ;)

I like their server farm. I've been to their colo facility. Pretty impressive.

[ab0wr]

ki6nno:
However, air and every other material in the universe does have some amount of conductivity and thus some amount of charge will migrate beyond the surface of a conductor into whatever medium is surrounding it. Going back to the point charge example, anytime a difference in charge exists between two points, an electrical field exists between those points as well.

Wasn't it you that pointed out that an isolated conductor could NOT have a net electrostatic charge that could be measured by an electroscope?

If what you say above is true then an isolated conductor that has* been* carrying current *could* have a remnant positive charge because it has lost negative charge (i.e. electrons) to the surrounding medium.

It would seem that one of these statements must not be correct.

My engineer training did not equate electric flux between two point charges to the E-field associated with a propagating E-M field. It would seem that you are trying to make the two the same. Is that correct?

tim ab0wr

[W4DFW]

I'm still flipped out that Quantum Mechanics and Newtonian Physics are at odds . . . . . and now THIS??

GAWD!!! :mad:

[KI6NNO]

Hi Tim,

Wasn't it you that pointed out that (1) an isolated conductor could NOT have a net electrostatic charge that could be measured by an electroscope?

If what you say above is true then (2) an isolated conductor that has* been* carrying current *could* have a remnant positive charge because it has lost negative charge (i.e. electrons) to the surrounding medium.

(3) It would seem that one of these statements must not be correct.

(numbers/color/emphasis added by ki6nno for clarity)

1) Yes, 2) No - not if it has the same total charge it started with at the time of measurement, 3) No. I believe that the issue is oversimplification of what I'm trying to say.

1) An isolated conductor cannot have a net electrostatic charge (differentially across its length/width/height) that can be measured by an electroscope - primarily because the electroscope isn't sensitive enough to measure extremely small differences in charge that if/when they occur, happen faster than the leaves are able to move. I didn't re-read the threads, so hopefully I answered this within the original context I wrote the first response in.

Beyond that, conservation of charge dictates that if an isolated conductor were to have a differential charge across it, any difference in charge must come from within the conductor itself (assuming we started with exactly the charge that was "in" the conductor to begin with and there are no external effects - which is not possible). However, quantum effects in QED allow the probability that a given charge can exist at any point on the conductor, which would allow for differential charges to exist for very short periods of time within the conductor, while the net charge on the isolated conductor would remain constant. This again goes back to Heisenberg's uncertainty principle in (not) being able to determine exactly what charge is located where and at what time.

2) If there is no "pressure" (charge differential) to cause an attractive or repulsive force between charges, the charges will not migrate. Thus an isolated conductor with the same charge as its surroundings (grounded or whatever) will not gain or lose charge unless something external to the isolated conductor causes a pertubation of the system, and causes a differential charge to appear on the conductor in relation to its environment.

A conductor that has been "charged" by contacting a charged object like a Van der Graaf generator, will remain charged when it is removed from contact with the charged object until the charges on it migrate (discharge) into its surrounding environment be it air, another object, or vacuum (like in a tube - even a vacuum has permittivity and conductivity). At some point, the net charge on the conductor will become equivalent to that of its environment and no further charge migration will occur.

In the electrolytic capacitor example, charges are able to migrate into the capacitor electrolyte due to the differential charge induced by the supply powering the circuit. Now, when the capacitor is discharged and the power supply disconnected, excess charge in the electrolyte will discharge from the electrolyte to the electrode plates until the forces between the charges reach a "ground state" or minimum force. One plate will become somewhat more positively charged than it was immediately after the capacitor was discharged, and the other will become somewhat more negatively charged. However, the net charge in the entire capacitor will be exactly the same as it was immediately after the capacitor was discharged and removed from the circuit. Over time (which can be a looooong time, if the net charge in the capacitor is greater than that of its environment, the excess charge in the "discharged and rested" capacitor will migrate from the electrolyte and plates in to the capacitor's environment until such time as the entire capacitor system reaches a charge ground state with its environment.

My engineer training did not equate electric flux between two point charges to the E-field associated with a propagating E-M field. It would seem that you are trying to make the two the same. Is that correct?

Yes and no (sorry, best I can do).
Einstein (with due credit to Maxwell, Faraday, and Heaviside) already unified the electric and magnetic forces in Special Relativity. I'm just trying to make a point about it in lay terms. I can't possibly do the math to prove it this far out of college, and only then because I was shown how it was done.

Now, a magnetic field is created when an e-field is in motion from the perspective of the observer. This is why sometimes an E-field is an E-field and sometimes it's an E-M field. But also, a changing E-field causes a change in the M-field (H) and vice versa. This latter observation is what makes a radio wave self-propagating. From an AC standpoint, any collapsing E-field will cause an expanding M-field. The permittivity (permeability, etc.) of the material within the system this is happening in determines where the vast majority of the electric and magnetic field lines will flow. If a material is conductive, an electric current will flow through it, if permittive, electric field lines will tend to flow through it, and if permeable, magnetic lines. With DC, there is motion of the E-field in the conductor and thus a magnetic field will form. In a constant electrostatic system however, there is no (or trivial) E-field movement and a magnetic field does not form (beyond what is caused by trivial charge migration into the environment). Even if there is charge migration into the environment, the migration must be directional enough for the small magnetic fields created by each charge's migration vector to "sum up" into a large enough magnetic field to be measureable.

So what I'm saying is that there are observational constraints (relative e-field motion) that determine whether or not a magnetic field is present, as well as dynamic requirements that determine both whether an E and/or H field is present.

Sorry Tim, I'm past the point I can think on this further this morning. It's past 2am and I hope I haven't introduced any inconsistancies.

73, Dave.

[ab0wr]

Hi Tim,

1) An isolated conductor cannot have a net electrostatic charge (differentially across its length/width/height) that can be measured by an electroscope - primarily because the electroscope isn't sensitive enough to measure extremely small differences in charge that if/when they occur, happen faster than the leaves are able to move. I didn't re-read the threads, so hopefully I answered this within the original context I wrote the first response in.

I'm sorry but this doesn't sound correct either. Why is the electrostatic charge on an isolated conductor changing so fast that the electroscope leaves can't move fast enough? What is causing this to happen in an *isolated* conductor?

Besides, the issue isn't about the differential charge from one end of the conductor to the other but instead the total charge of the entire conductor compared to the rest of the environment.

If a conductor "boils" off electrons into the "ether" while it is carrying current, it *wlll* wind up with a net positive charge when the circuit is opened and the voltage generator is removed. That net positive charge would be distributed equally throughout the conductor.

Beyond that, conservation of charge dictates that if an isolated conductor were to have a differential charge across it, any difference in charge must come from within the conductor itself (assuming we started with exactly the charge that was "in" the conductor to begin with and there are no external effects - which is not possible). However, quantum effects in QED allow the probability that a given charge can exist at any point on the conductor, which would allow for differential charges to exist for very short periods of time within the conductor, while the net charge on the isolated conductor would remain constant. This again goes back to Heisenberg's uncertainty principle in (not) being able to determine exactly what charge is located where and at what time.

I think you just said what I pointed out above that your conclusions lead to, the isolated conductor *can* have a net overall charge just not a *differential* charge.

2) If there is no "pressure" (charge differential) to cause an attractive or repulsive force between charges, the charges will not migrate. Thus an isolated conductor with the same charge as its surroundings (grounded or whatever) will not gain or lose charge unless something external to the isolated conductor causes a pertubation of the system, and causes a differential charge to appear on the conductor in relation to its environment.

A conductor that has been "charged" by contacting a charged object like a Van der Graaf generator, will remain charged when it is removed from contact with the charged object until the charges on it migrate (discharge) into its surrounding environment be it air, another object, or vacuum (like in a tube - even a vacuum has permittivity and conductivity). At some point, the net charge on the conductor will become equivalent to that of its environment and no further charge migration will occur.

Yet I've *never* heard of a conductor being "charged" from carrying current. Have you?


Now, a magnetic field is created when an e-field is in motion from the perspective of the observer. This is why sometimes an E-field is an E-field and sometimes it's an E-M field. But also, a changing E-field causes a change in the M-field (H) and vice versa.

But the question was: is the electric flux field between two charges the same as the E-field generated by a moving charge?

You didn't answer the question.

tim ab0wr

[KI6NNO]

Well, that's what I get for trying to do this at 2am - my fault completely. :eek:

I took another stab at it in-between activities at work. I hope this came out better. It's still pretty lengthy so I'll try and pare it down a bit later on.

I'm sorry but this doesn't sound correct either. Why is the electrostatic charge on an isolated conductor changing so fast that the electroscope leaves can't move fast enough? What is causing this to happen in an *isolated* conductor?

I never said the net charge on an isolated conductor is changing so fast on an isolated conductor that the electroscope leaves can't move fast enough. I was trying to convey that quantum electrodynamics allows a charge at a point in or on a conductor to exist or not exist at that point with a fair amount of unpredictability however conservation of charge requires that the net charge remains the same. Therefore if a charge "appears" at a location on an isolated conductor, an equivalent charge must "disappear" at some other location to satisfy the conservation of charge criterion.

Here's a conductor with four points (atoms) that for illustrative sake is completely isolated from the universe:

==A(0)==B(0)==C(0)==D(0)==

Since metal atoms are neutral, there is no net charge on the conductor.
However QED allows for the probability of charge at a local point to change, so the conductor could look like this for a brief time:

==A(-)==B(0)==C(0)==D(+)==

While this would create a small differential charge in the conductor, the net charge is the conductor is still 0. Further more, there's a great deal of force created by the charge differential and nature won't allow things to stay like that for very long. So perhaps the conductor now looks like this:

==A(0)==B(+)==C(-)==D(0)==

again, net zero charge, but with a small differential charge due to QED.

This can't be measured by an electroscope because the time scale for QED effects is much to short for any macroscopic physical effect to be imparted on the electroscope leaves. (Fwiw, all of the electroscope's physical components are undergoing similar QED effects all the time, again with tiny localized differential charges coming and going in very short time scales with no physical effects).

Since the charges are moving, there is a small local magnetic field created. However, since the net charge is 0 and there is no directional current flow for any detectable amount of time, there is no measureable "net" magnetic field in the conductor.

Besides, the issue isn't about the differential charge from one end of the conductor to the other but instead the total charge of the entire conductor compared to the rest of the environment.

In an electrically neutral conductor (net zero charge) in an electrically neutral environment - say a tungsten filament in a disconnected conventional light bulb, there is no charge differential between the conductor (filament) and vacuum and no charge migration save for QED effects.

In an electrostatically charged conductor (let's say negatively charged for illustrative sake), charges will migrate back into the surrounding environment because there is a relative deficit of negative charges outside of the conductor. This is because the negative charges that make up the excess negative charge of the conductor came from the surrounding environment. They did not somehow pop into existance in the conductor and stay there. It is the permittivity and conductivity of the conductor and the material and/or "space" of the conductor's environment that allows E-fields to flow and charge to migrate giving the charge differential between the conductor and environment a path to "discharge" and reach a neutral "ground state" condition.

If a conductor "boils" off electrons into the "ether" while it is carrying current, it *wlll* wind up with a net positive charge when the circuit is opened and the voltage generator is removed. That net positive charge would be distributed equally throughout the conductor.
No. the electrons that entered the conductor and were "boiled off" into the environment came from the power source (environment) and not the conductor. When all discharging is completed, the conductor will have exactly the same number of electrons it started with and will be electrically (net) neutral to itself and (net) neutral with its environment.

Beyond that, conservation of charge dictates that if an isolated conductor were to have a differential charge across it, any difference in charge must come from within the conductor itself (assuming we started with exactly the charge that was "in" the conductor to begin with and there are no external effects - which is not possible). However, quantum effects in QED allow the probability that a given charge can exist at any point on the conductor, which would allow for differential charges to exist for very short periods of time within the conductor, while the net charge on the isolated conductor would remain constant. This again goes back to Heisenberg's uncertainty principle in (not) being able to determine exactly what charge is located where and at what time.
I think you just said what I pointed out above that your conclusions lead to, the isolated conductor *can* have a net overall charge just not a *differential* charge. Yet I've *never* heard of a conductor being "charged" from carrying current. Have you?

Not exactly. As noted above a conductor does not gain or lose charge from carrying current.

A conductor can gain or lose charge by becoming electrostatically charged (as in an open circuit with voltage applied to it or for example from a Van der Graaf generator). But even an open circuit isn't exactly "open" because of the permittivity and conductivity properties of everything in the universe including the ether. So, yes a conductor can have a net overall positive or negative charge from being electrostatically charged. It will not stay that way for long.

Example:
Let's say we have a Van der Graaf generator and a piece of wire is touched to the negatively charged sphere. The wire will gain negative charge from the sphere. If the wire is removed, the negative charges must discharge back into the environment because there is now a differential charge between the conductor and the surrounding environment. Also, the atoms in the conductor will not accept free electrons as they want to become electrically neutral and reach their own ground states. Therefore there is a great deal of physical force (on an atomic scale) in the conductor trying to get rid of the excess charge from the conductor.

Now, a magnetic field is created when an e-field is in motion from the perspective of the observer. This is why sometimes an E-field is an E-field and sometimes it's an E-M field. But also, a changing E-field causes a change in the M-field (H) and vice versa.

But the question was: is the electric flux field between two charges the same as the E-field generated by a moving charge?
You didn't answer the question.

The term "electric flux field" isn't really a useable term. I think I know what you mean, and that might be the disjuct here.
The electric flux is by definition the strength of the electric field multiplied by the area being measured that's perpendicular to the electric field.
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gaulaw.html#c3

or: Flux = EA cos Theta

Now an electric field is an electric field. It doesn't matter if the charges are stationary or moving relative to each other or relative to an observer.
So to answer your question as I understand it: yes, the "electric flux field" between two charges is the same as the E-field generated by (two) moving charge(s).

But wait, there's more - the answer isn't quite that simple except in the most trivial case.

Describing the field is an entirely different exercise because there must be sufficient charge distribution symmetry and physical symmetry to be able to apply simple geometry to the object(s) being measured for Gauss's law to work the way it's written (above).

If there isn't enough symmetry in the object or the charge distribution of the object to apply Gauss's law directly to an object, the fields generated from charges on "points" on the object(s) must be summed in order to calculate the E-field strength.

Also, Gauss' law includes the angle of the surface relative to the direction of the field. Thus, if Theta changes over time due to motion, then the electrical flux must change as well.

So, the second part of the answer is:
For moving charges, the electric flux will change because Theta (the angle of the area perpendicular to the electric field) will change as the charges move relative to each other. This also holds true for a charge or charges that are moving relative to the observer.

(6/12) That still doesn't answer your question because once there's an alternating electric and magnetic field, the "wave" leaving the vicinity of the charges is self-propagating. It doesn't need charges anymore to create an electric field, and the charges don't have to move to create the magnetic field - they essentially reinforce each other forever although the strength of the wave diminishes with distance from its point of origin.
Hopefully that takes care of answering it. :)

73, Dave